A bcc metal (M=56 g mol^-1) has atomic radius 125 pm. Use N_A = 6.022x10^23 mol^-1. Using sqrt3=1.732, calculate its density in g cm^-3 (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss162

A bcc metal ( $M = 56 g mo l^{- 1}$ ) has atomic radius $125 pm$ . Use $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ . Using $3 = 1.732$ , calculate its density in $g c m^{- 3}$ (two decimals).

Numerical answer type, answer in g/cm^3, tolerance 0.03.

For bcc,

a = \frac{4 r}{3} = \frac{4 \times 125}{1.732} = 288.7 pm

and

Z = 2

ρ = \frac{2 \times 56}{6.022 \times 1 0 ^{23} \times ( 2.887 \times 1 0 ^{- 8} ) ^{3}} = 7.73 g c m^{- 3}